Problem: $h(t) = -6t+1$ $f(x) = 2+4(h(x))$ $g(t) = 3t^{2}+4t-6-f(t)$ $ h(f(2)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(2)$ . Then we'll know what to plug into the outer function. $f(2) = 2+4(h(2))$ To solve for the value of $f$ , we need to solve for the value of $h(2)$ $h(2) = (-6)(2)+1$ $h(2) = -11$ That means $f(2) = 2+(4)(-11)$ $f(2) = -42$ Now we know that $f(2) = -42$ . Let's solve for $h(f(2))$ , which is $h(-42)$ $h(-42) = (-6)(-42)+1$ $h(-42) = 253$